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Question Number 163496 by Ahmed777hamouda last updated on 07/Jan/22

 ∫_0 ^(π/2) 256cos^5 ((x/2))sin^(11) ((x/2))dx

0π2256cos5(x2)sin11(x2)dx

Answered by Ar Brandon last updated on 07/Jan/22

I=∫_0 ^(π/2) 256cos^5 ((x/2))sin^(11) ((x/2))dx=8∫_0 ^(π/2) sin^5 x(((1−cosx)/2))^3 dx     =∫_0 ^(π/2) sin^5 x(1−3cosx+3cos^2 x−cos^3 x)dx     =(1/2)β(3, (1/2))−(3/2)β(3, 1)+(3/2)β(3, (3/2))−(1/2)β(3, 2)     =(2/5)∙(2/3)∙(2/1)−(1/2)+(3/2)∙(2/7)∙(2/5)∙(2/3)∙(2/1)−(1/(4×3×2×1))     =(8/(15))−(1/2)+(8/(35))−(1/(24))=((37)/(168))

I=0π2256cos5(x2)sin11(x2)dx=80π2sin5x(1cosx2)3dx=0π2sin5x(13cosx+3cos2xcos3x)dx=12β(3,12)32β(3,1)+32β(3,32)12β(3,2)=25232112+322725232114×3×2×1=81512+835124=37168

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